Estude! Com Prof. Caju

Calcule o valor da expressão E = [tex3]\frac{\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}+1}}}{\sqrt{\sqrt[4]{8}+\sqrt{\sqrt{2}-1}}-\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}-1}}}[/tex3]

@FISMAQUIM,

[tex3]a = \sqrt[4]{8} \\b = \sqrt{\sqrt{2}+1}\\c = \sqrt{\sqrt{2}-1}\\
X = \frac{\sqrt{a-b}}{\sqrt{a+c} - \sqrt{a-c}}[/tex3].
Analisando o Denominador: Seja [tex3]D = \sqrt{a+c} - \sqrt{a-c}[/tex3]
[tex3]D^2 = (\sqrt{a+c})^2 + (\sqrt{a-c})^2 - 2\sqrt{(a+c)(a-c)} = (a+c) + (a-c) - 2\sqrt{a^2 - c^2}\\D^2 = 2a - 2\sqrt{a^2 - c^2}.[/tex3]
[tex3]a^2 = (\sqrt[4]{8})^2 = \sqrt{8} = 2\sqrt{2}\\c^2 = (\sqrt{\sqrt{2}-1})^2 = \sqrt{2}-1[/tex3]
Substituindo: [tex3] a^2 - c^2 = 2\sqrt{2} - (\sqrt{2}-1) = \sqrt{2} + 1[/tex3]
[tex3]b^2 = \sqrt{2} + 1. \therefore \sqrt{a^2 - c^2} = \sqrt{\sqrt{2} + 1} = b [/tex3]
Substituindo de volta em [tex3]D^2: D^2 = 2a - 2b = 2(a-b)[/tex3]
[tex3]X^2 = \frac{(\sqrt{a-b})^2}{D^2} = \frac{a-b}{2(a-b)} \implies X^2 = \frac{1}{2}[/tex3]
[tex3] X = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} =\boxed{ \frac{\sqrt{2}}{2}}[/tex3]